Question 1: Find the perimeter and area of a rectangle whose length and breadth are 20 \ cm and 8 \ cm respectively.

Answer:

Dimensions of the rectangle: Length (l) = 20 \ cm      Breadth (b) = 8 \ cm

Therefore Perimeter = 2 (l + b) = 2 ( 20 + 8) = 56 \ cm

Area = l \times b = 20 \times 8 = 160 \ cm^2

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Question 2: A rectangular room floor is 192 \ m^2 in area. If its length is 16 \ m , find its perimeter.

Answer:

Dimensions of the rectangle: Length (l) = 16 \ m    Let Breadth (b) = x \ m

Area of rectangle = l \times b

\Rightarrow 192 = 16 \times x \Rightarrow  x = 12

Hence Breadth = 12 \ m

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Question 3: Find the length of a diagonal of a rectangle whose adjacent sides are 8 \ m and 6 \ m long.

Answer:

Dimensions of the rectangle: Length (l) = 8 \ m    Let Breadth (b) = 6 \ m

Diagonal of a rectangle = \sqrt{l^2 + b^2} =  \sqrt{8^2 + 6^2} = \sqrt{100} = 10 \ m

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Question 4: Find the length of a diagonal of a square of side 4 \ cm .

Answer:

Dimension of a square: Side (a) = 4 \ cm

Diagonal of a square = \sqrt{2} a = 4 \sqrt{2} \ cm

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Question 5: Find the perimeter of a square the sum of the lengths of whose diagonal is 144 \ cm .

Answer:2019-01-20_20-04-08

Dimension of a square: Side = a

Given: Diagonal of the square = 144 \ cm

We know diagonal of a square = \sqrt{2} a

\Rightarrow 2 (\sqrt{2} a) = 144

\Rightarrow a = \frac{72}{\sqrt{2}} = 36\sqrt{2} \ cm

Perimeter of a square = 4a = 4 \times 36\sqrt{2} = 144\sqrt{2} \ cm

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Question 6: The length and breadth of a room are in the ratio 3 : 2 . Its area is 216 \ m^2 . Find its perimeter.

Answer:

Dimensions of the rectangle: Let Length (l) = 3x    Let Breadth (b) = 2x

Given: Area is 216 \ m^2

Therefore 216 = 3x \times 2x

\Rightarrow x^2 = 36 

\Rightarrow x = 6

Hence  Length (l) = 18 \ m    Let Breadth (b) = 12 \ m

Therefore Perimeter = 2 (l + b) = 2 (18 + 12) = 60 \ cm

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Question 7: The diagonal of a square A is (a + b) . Find the diagonal of a square B whose area is twice the area of A .

Answer:

Given: Diagonal of a square A is (a + b)

If the side of the square = x

\Rightarrow Diagonal = \sqrt{2} x = (a+b)

\Rightarrow x = \frac{a+b}{\sqrt{2}}    … … … … … (i)

Let the side of the second square = y

Given: y^2 = 2 x^2 \Rightarrow y = \sqrt{2} x

Diagonal = \sqrt{2} y = \sqrt{2} \times \sqrt{2} x = 2x    … … … … … (ii)

Substituting (i) in (ii)  we get

Diagonal = 2 \times \Big( \frac{a+b}{\sqrt{2}} \Big) = \sqrt{2} (a+b)

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Question 8: The perimeter of a square is (4x + 20) \ cm . Find its diagonal.

Answer:

Perimeter = 4x + 20

\Rightarrow Side = \frac{4x+20}{4} = x+5

Hence Diagonal = \sqrt{2} (x+5)

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Question 9: Find the area of a square that can be inscribed in a circle of radius 10 \ cm .2019-01-20_20-16-12

Answer:

Radius = 10 \ cm

Hence the length of the side = \sqrt{10^2 + 10^2} = 10 \sqrt{2}

Therefore Area = 10 \sqrt{2} \times 10 \sqrt{2} = 200 \ cm^2

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Question 10: Find the perimeter of a square the sum of the lengths of whose diagonals is 100 \ cm .

Answer:2019-01-20_20-15-31

Let the side be = x

Given: \sqrt{2} x + \sqrt{2} x = 100

\Rightarrow 2 \sqrt{2} x = 100

\Rightarrow x = \frac{50}{\sqrt{2}} = 25 \sqrt{2}

Therefore Perimeter = 4 \times 25 \sqrt{2} = 100 \sqrt{2} \ cm

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Question 11: The diagonal of a square is 14 \ cm . Find its area.

Answer:

Let the side of the square = x

Given: \sqrt{2} x = 14

\Rightarrow x = 7 \sqrt{2}

Therefore Area = 7 \sqrt{2} \times 7 \sqrt{2} = 28 \ cm^2

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Question 12: Find the area and perimeter of a square plot of land the length of whose diagonal is 15 \ m .

Answer:

Given: Diagonal = 15 \ m

Let the side of the square = x

Therefore \sqrt{2} x = 15

\Rightarrow x = \frac{15}{\sqrt{2}} 

Therefore Area = \frac{15}{\sqrt{2}} \times \frac{15}{\sqrt{2}} = \frac{225}{2} = 112.5 \ m^2

Hence the Perimeter = 5 \times \frac{15}{\sqrt{2}} = 30 \sqrt{2}

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Question 13: Find the ratio of the area of a square to that of the square drawn on its diagonal.

Answer:

Let the side of square = x

Therefore diagonal = \sqrt{2} x

Hence the ratio = \frac{x^2}{(\sqrt{2} x)^2} = \frac{1}{2} 

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Question 14: The diagonal of square A is (a + b) . Find the diagonal of square B whose area is half of the area of A .

Answer:

Let the side of square A = x

Therefore \sqrt{2} x = a+b

\Rightarrow x = \frac{a+b}{\sqrt{2}} 

Let the side of square B = y

Therefore y^2 = \frac{1}{2} x^2

\Rightarrow y = \frac{1}{\sqrt{2}} x

Diagonal of square B = \sqrt{2} y = \sqrt{2} \times \frac{1}{\sqrt{2}} x = \frac{a+b}{\sqrt{2}}

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Question 15: The perimeter of a square is 48 \ m . The area of a rectangle is 4 \ sq. m less than the area of the given square. If the length of the rectangle is 14 \ m , find its breadth.

Answer:

Let the side of the square = a

Therefore 4a = 48 \Rightarrow a = 12 \ m

Let the breadth of the rectangle = b

Therefore 14 \times b = 12^2 - 4

\Rightarrow 14b = 144 - 4

\Rightarrow b = 10 \ m

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Question 16: The perimeter of one square is 748 \ cm and that of another is 336 \ cm . Find the perimeter and the diagonal of a square whose area is equal to the sum of the areas of these two squares.

Answer:

Let the side of square 1 = a_1

Therefore 4 a_1 = 748 \Rightarrow a_1 = 187 \ cm

Hence the area of square 1 = 187^2 = 34969 \ cm^2

Let the side of square 2 = a_2

Therefore 4 a_2 = 336 \Rightarrow a_2 = 84 \ cm

Hence area of square 2 = 84^2 = 7056  \ cm^2

Therefore area of square 3 = 34969 + 7056 = 42025 \ cm^2

Hence the side of square 3 = \sqrt{42025} = 205 \ cm

Therefore the perimeter of square 3 = 4 \times 205 = 820 \ cm

Diagonal of square 3 = \sqrt{2} (205) = 289.91 \ cm

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Question 17: The perimeter of a rectangular card board is 96 \ cm . If its breadth is 18 \ cm , find the length and area of the card board.

Answer:

Dimensions of the rectangle: Length (l) = l      Breadth (b) = 18 \ cm

Therefore 2(l+b) = 96

\Rightarrow l+18 = 48

\Rightarrow l = 30 \ cm

Hence area = 30 \times 18 = 540 \ cm^2

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Question 18: If the sides of two squares are in the ratio x : y , prove that their areas are in the ratio x^2:y^2 .

Answer:

Side of square 1 = x

Therefore Area of square 1 = x^2

Side of square 2 = y

Therefore Area of square 2 = y^2

Therefore ratio of areas = \frac{x^2}{y^2} 

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Question 19: In exchange for a square plot one of whose sides is 84 \ m , a man wants to buy a rectangular plot 144 \ m long and of the same area as of the square plot. Find the width of the rectangular plot.

Answer:

Side of square plot = 84 \ m

Length of rectangular plot = 144 \ m

Let breadth of rectangular plot = b

Therefore 144 \times b= 84 \times 84

\Rightarrow b = 49 \ m

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Question 20: A rectangular lawn 80 \ m \times 60 \ m has two roads each with 10 \ m wide running in the middle of it, one parallel to the length and other parallel to the breadth. Find the cost of graveling them at 30 paisa per square meter.2019-01-20_20-15-40

Answer:

Area of graveled road = 10 \times 80 + 10 \times 60 - 10 \times 10 = 800 + 600 - 100 = 1300 \ m^2

Therefore cost of graveling = 1300 \times 0.30 = 390 \ Rs.

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Question 21: The area of a square plot is \frac{1}{2} hectare. Find the diagonal of the square.

Answer:

Let the side of square plot = a

Therefore a^2 = \frac{1}{2} hectare

We know 1 hectare = 10000 \ m^2

Therefore a^2 = 5000 \Rightarrow a = 50 \sqrt{2}

Hence diagonal = \sqrt{2} a = \sqrt{2} (50 \sqrt{2}) = 100 \ m

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Question 22: A lawn is in the form of a rectangle having its sides in the ratio 5: 2 . The area of the lawn is 1000 \ m^2 . Find the cost of fencing it at the rate of Rs. \ 8.50 per meter.2019-01-20_20-15-18

Answer:

Let length = 5x and breadth = 2x

Therefore 5x \times 2x = 1000 \Rightarrow 10x^2 = 1000 \Rightarrow x = 10 \ m

Therefore length = 50 \ m and breadth = 20 \ m

Therefore perimeter = 2 (50 + 20) = 140 \ m

Therefore cost of fencing = 140 \times 8.5 = 1190 \ Rs.

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Question 23: The area of a square park is 40,000 \ sq. m . Find the cost of fencing it at the rate of Rs. \ 2.80 per meter.

Answer:

Area of square = 40000 \ m^2

Let side of the square = a

\Rightarrow a^2 = 40000 \Rightarrow a = 200 \ m

Therefore Perimeter = 4a = 800 \ m

Therefore cost of fencing = 800 \times 2.8 = 2240 \ Rs.

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Question 24: The area of the base of a rectangular tank is 2400 \ m^2 and its sides are in the ratio 3: 2 . Find the cost of planting flowers round it at the rate of Rs. \ 1.25 per meter.

Answer:

Let length = 3x and breadth = 2x

Therefore 3x \times 2x = 2400 \Rightarrow x^2 = 400 \Rightarrow x = 20 \ m

Therefore length = 60 \ m and breadth = 40 \ m

Therefore perimeter =2 (60 + 40) = 200 \ m

Therefore cost of planting flowers = 200 \times 1.25 = 250 \ Rs.

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Question 25: A rectangular field 242 meter long has got an area of 4840 \ sq. m , what will be the cost of fencing that field on all the four sides, if 1 meter of fencing costs 20 paisa?

Answer:

Length = 242 \ m

Area = 4840 \ m^2

Therefore breadth = \frac{4840}{242} = 20 \ m

Perimeter = 2 (242 + 20) = 524 \ m

Therefore cost of fencing = 524 \times 0.2 = 1048 \ Rs.

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Question 26: A rectangular grassy plot is 112 \ m \times 78 \ m . It has gravel path 2.5 \ m wide all around it on the inside. Find the area of the path and the cost of constructing it at the rate of Rs. \ 3.40 per sq. meter.

Answer:

Dimensions of park: Length = 112 \ m , Breadth = 78 \ m

Inner Dimensions of park: Length = (112 - 5) = 107 \ m , Breadth = (78-5) = 73  \ m

Area of path = 112 \times 78 - 107 \times 73 = 925 \ m^2

Therefore cost of construction = 925 \times 3.4 = 3145 \ Rs.

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Question 27: There is a square field whose side is 44 \ m . A flowerbed is prepared in its center, leaving a gravel path of uniform width all around the flower bed. The total cost of laying the flower bed and graveling the path at Rs. \ 2 and Rs. \ 1 per square meter respectively is Rs. \ 3536 . Find the width of the gravel path.

Answer:2019-01-20_20-23-20

Let the side of the garden = x

Therefore the area of the garden = x^2

Area of the path = 44^2 - x^2

Therefore 3536 = 2 \times x^2 + 1 \times ( 44^2 - x^2)

\Rightarrow 3536 = 2x^2 + 44^2 - x^2

\Rightarrow x^2 = 3536 - 44^2 = 1600

\Rightarrow x = 40 \ m

Hence the path = 2 \ m \ wide

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Question 28: How many tiles 40 \times 40 \ cm^2 each will be required to pave the footpath 1 \ m wide carried round the outside of a grassy plot 28 \ m by 18 \ m ?2019-01-20_20-27-10

Answer:

Are of footpath = 20 \times 30 - 28 \times 18 = 96 \ m^2

No of times required = \frac{96}{0.4 \times 0.4} = 600 \ tiles

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Question 29: A room is 7.4 \ m long, 3.6 \ m broad and 4 \ m high. It has two doors 2.1 \ m \times 1.2 \ m and 5 windows each 1.8 \ m \times 1.2 \ m . How much will it cost to whitewash the walls of the room at the rate of Rs.\ 2.00 per square meter?

Answer:

Dimensions of the Room: Length (l) = 7.4 \ m , Breadth (b) = 3.6 \ m and Height (h) = 4 \ m

Dimensions of the Doors: Breadth (b) = 2.1 \ m and Height (h) = 1.2 \ m

Dimensions of the Window: Breadth (b) = 1.8 \ m and Height (h) = 1.2 \ m

Area of walls = Lateral Area = 2 (l+b) h = 2 (7.4 + 3.6) \times 4 = 88 \ m^2

Area of Doors and Windows = 2 \times 2.1 \times 1.2 + 5 \times 1.8 \times 1.2 = 15.84 \ m^2

Area to be painted = 72.16 \ m^2

Cost of painting = 72.16 \times 2 = 144.32  \ Rs.

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Question 30: A carpet is laid on the floor of a room 8 \ m \times 5 \ m . There is a border of constant width around the carpet. If the area of the border is 1.2 \ m^2 , find its width.

Answer:

Dimensions of the Room: Length (l) = 8 \ m , Breadth (b) = 5 \ m

Area of Border = 14 \ m^2

Therefore 14 = (5+2x)(8+2x) - 8 \times 5

\Rightarrow 12 = 40 + 16x + 10x +4x^2 - 40

\Rightarrow 4x^2 + 26x^ - 14 = 0

\Rightarrow  2x^2 + 13x - 7 = 0

\Rightarrow (2x-1)(x+7) = 0

\Rightarrow x = \frac{1}{2}  or x = -7 (this is not possible as x cannot be negative)

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Question 31: A rectangular courtyard, 3.78 \ m long and 5.25 \ m broad, is to be paved exactly -with square tiles, all of the same size. Find the largest size of such a tile and the number of tiles required to pave it.

Answer:

Dimension of courtyard: Length (l) = 3.78 \ m \ or \ 378 \ cm

Breadth (b) = 5.25 \ m \ or \ 525 \ cm

The largest common divisor of both the numbers is 21

Therefore the dimension of the largest square tile is 21 \ cm

Hence the number of tiles required = \frac{3.78 \times 5.25}{0.21 \times 0.21} = 450

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Question 32: The cost of fencing a square field at 60 paisa per meter is Rs. \ 1200 . Find the cost of reaping the field, it the rate of 50 \ paisa per 100 \ sq. m .

Answer:

Perimeter \times 0.60 = 1200

\Rightarrow Perimeter = 2000 \ m

Therefore side = \frac{2000}{4} = 500 \ m

Therefore Area = 500 \times 500 = 250000

Therefore cost of reaping the field = \frac{0.50}{100} \times 250000 = 1250 \ Rs.

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Question 33: A room 4.9 \ m long and 3.5 \ m broad is carpeted with a carpet, leaving an uncovered margin 25 \ cm all around the room. If the breadth of the carpet is 80 \ cm , find its cost at Rs. \ 60 per meter.

Answer:

Dimensions of the Room: Length (l) = 4.9 \ m , Breadth (b) = 3.5 \ m

Dimensions of the Carpet: Length (l) = (4.9-0.5) = 4.4 \ m , Breadth (b) = (3.5 - 0.5) = 3.0  \ m

Area = 4.4 \times 3 = 13.2 \ m^2

Let the length of the carpet needed = l

Therefore l \times 0.80 = 13.2

\Rightarrow l = \frac{13.2}{0.80} = 16.5 \ m

Therefore cost = 16.5 \times 0.60 = 9.9 \ Rs.

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Question 34: The cost of carpeting a room at Rs. \ 2.5 Per square meter is Rs. \ 450 . The cost of whitewashing the walls at 50 paisa per sq. meter is Rs. \ 135 . The room is 12 \ m wide. Find its height.

Answer:

12 \times b \times 2.5 = 450

\Rightarrow b = \frac{450}{12 \times 2.5} = 15 \ m

Cost of paining area of walls

2 (l+b) h \times 0.5 = 135

\Rightarrow 2(12 + 15) h \times 0.5 = 135

\Rightarrow h = \frac{135}{27} = 5

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Question 35: A room 5\ m long and 4 \ m wide is surrounded by a verandah. Find the width of the verandah if it occupies 22 square meters.

Answer:

Let x be the width of the varanda

Therefore (5+2x)(4+2x) - 20 = 22

\Rightarrow 20 + 8x + 10x + 4x^2 - 20 = 22

\Rightarrow 4x^2 + 18x - 22 = 0

\Rightarrow (2x-2)(x +11) = 0

\Rightarrow x = 1 or x = -11 (not possible)

Therefore the width of the verandah = 1 \ m

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Question 36: Square carpet is spread in the center of a room 55 \ dm square leaving a small margin of equal width all around. The total cost of carpeting at 25 \ paisa \ per \ square \ dm and decorating the margin at 15 \ paisa \ per \ square \ dm is Rs. \ 703.75 . Find the width of the margin.

Answer:

Dimension of square room = 55 dm

Let the width of the margin = a

Therefore area of carpet = (55 - 2a)^2

Area of margin = 55^2 - (55 - 2a)^2

\Rightarrow 0.4a^2 - 22a + 52.5 = 0

\Rightarrow a^2 - 55a + 131.25 = 0

\Rightarrow (a-2.5)(a-52.5) = 0

\Rightarrow a = 2.5  \ or \ 52.5 (not possible)

Hence the width of the margin = 2.5 \ dm

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Question 37: If the length and breadth of a room are increased by 1 \ m the area is increased By 21 \ m^2 . If the length is increased by 1 \ m and breadth is decreased by 1\ m , the area is decreased by 5 \ m^2 . Find the perimeter of the room.

Answer:

(l+1)(b+1) - lb = 21 … … … … … i)

lb - (l+1)(b-1) = 5 … … … … … ii)

From i) lb + b + l +1 - lb = 21

\Rightarrow l+b = 20 … … … … … iii)

Therefore Perimeter = 2(l+b) = 2 \times 20 = 40 \ m

from ii) lb - (lb + b -l-1) = 5

\Rightarrow -b + l = 4 … … … … … iv)

Solving (iii) and (iv) we get

l = 12 and b = 8 .

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Question 38: A rectangle has twice the area of a square. The length of the rectangle is 12 \ cm greater and the width is 8 \ cm greater than the side of the square. Find the perimeter of the square.

Answer:

Let the dimension of rectangle be l and b . And that of the square be a .

Therefore l = 12 + a and b = 8 +a

(12+a)(8+a) = 2a^2

\Rightarrow96 + 8a + 12a + a^2 = 2a^2

\Rightarrow96 + 20a = a^2

\Rightarrow a^2 - 20a-96 = 0

\Rightarrow(a-24)(a+4) = 0

\Rightarrow a = 24 \ cm

Therefore Perimeter of square = 4 \times 24 = 96 \ cm

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Question 39: If the perimeter of a rectangular plot is 58 \ m and the length of its diagonal is 26 \ m , find its area.

Answer:

Let the dimension of rectangle be l and b .

Therefore 2(l+b) = 68 \Rightarrow l+b = 34 … … … … … i)

\sqrt{l^2 + b^2} = 26 … … … … … ii)

(l+b)^2 = 1156

l^2 + b^2 +2 lb = 1156

Therefore l^2 + b^2 = 1156 - 2lb

Substituting in (ii)

\sqrt{1156-2lb} = 26

\Rightarrow 480 = 2lb \ or \ lb = 240

Hence the area is 240 \ cm^2

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Question 40: The length of a rectangular garden is 12 \ m more than its breadth. The numerical value of its area is equal to 4 times the numerical value of its perimeter. Find the dimensions of its garden.

Answer:

Area = (b+12)b

Perimeter = 2 (b+12+b) = 2(2b+12)

Therefore  (b+12)b = 4 \times 2 (2b+12)

\Rightarrow b^2 + 12b = 16b + 96

\Rightarrow b^2 - 4b - 96 = 0

\Rightarrow (b-12)(b+8) = 0

Therefore b = 12   or b = -8 (not possible)

Hence b = 12 . Therefore dimensions are 24 \ cm and 12 \ cm

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Question 41: A wire when bent in the form of an equilateral triangle encloses an area 36\sqrt{6} \ cm^2 of Find the area enclosed by the same wire when bent to form: (i) a square (ii) a rectangle whose length is 2 \ cm more than its width.

Answer:

h = \sqrt{a^2- (\frac{a}{2})^2} = \frac{\sqrt{3}}{2} a

\Rightarrow \frac{1}{2} \times a \times \frac{\sqrt{3}}{2} a = 36 \sqrt{3}

\Rightarrow a^2 = 4 \times 36

\Rightarrow a = 12 \ cm

Therefore Perimeter = 36 \ cm

i) When bent in a square

Side = \frac{36}{4} = 9 \ cm

Therefore area = 9 \times 9 = 8  \ cm^2

ii) Perimeter of rectangle = 36 \ cm

If Breadth = x

Then 2(x + 2+x) = 36

\Rightarrow 2x = 16 \Rightarrow x = 8 \ cm

Therefore dimensions are 10 \ cm and 8 \ cm

Hence Area = 10 \times 8 = 80 \ cm^2

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